Explanation 1) Concept: By using Bragg’s condition, we can find the allowed values of the order m . Then, we can find the wavelength for each m value for the shorter and longer wavelengths. 2) Formula: i) 2 d sin ⁡ θ = m λ ii) Wavelength range is from 95 p m to 140 p m 3) Given: i) θ = 45 ° ii) d = 275 p m 4) Calculations: The wavelengths must satisfy Bragg’s condition. 2 d sin ⁡ θ = m λ Thus, m λ = 2 ( 275 p m ) ( sin ⁡ 45 ° ) m λ = 389 p m m = 389 p m λ The range of wavelength is given from 95 p m to 140 p m . Therefore, For λ = 95 p m , m = 389 p m 95 p m = 4

11th Edition

ISBN: 9781119459132

Author: Halliday

Publisher: WILEY

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Question

Chapter 36, Problem 72P

To determine

To find:

a) The longest wavelength

b) Order number associated with the longest wavelength

c) Shortest wavelength

d) Order number associated with the of the shortest wavelength

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Chapter 36, Problem 71P

Chapter 36, Problem 73P

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Chapter 36 Solutions

FUNDAMENTALS OF PHYSICS (LLF)+WILEYPLUS

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a) The longest wavelength b) Order number associated with the longest wavelength c) Shortest wavelength d) Order number associated with the of the shortest wavelength

Solution Summary: The author explains how Bragg's condition can be used to find the wavelength for each m value for the shorter and longer wavelengths.

Concept explainers

To find: a) The longest wavelength b) Order number associated with the longest wavelength c) Shortest wavelength d) Order number associated with the of the shortest wavelength

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Chapter 36 Solutions

To find:

a) The longest wavelength

b) Order number associated with the longest wavelength

c) Shortest wavelength

d) Order number associated with the of the shortest wavelength